You have found the following ages (in years) of all 4 zebras at your local zoo: $ 7,\enspace 20,\enspace 1,\enspace 2$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 20 + 1 + 2}{{4}} = {7.5\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-0.5$ years $0.25$ years $^2$ $20$ years $12.5$ years $156.25$ years $^2$ $1$ year $-6.5$ years $42.25$ years $^2$ $2$ years $-5.5$ years $30.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.25} + {156.25} + {42.25} + {30.25}} {{4}} $ $ {\sigma^2} = \dfrac{{229}}{{4}} = {57.25\text{ years}^2} $ The average zebra at the zoo is 7.5 years old. The population variance is 57.25 years $^2$.